Given a matrix A over field F of order 5x5, with tr(A)=0, and p(A)=1, find the eigenvalues of A.
If the question would have been over C/R I would have found an answer, but this F field thing is too much for me...
Thanks!
Given a matrix A over field F of order 5x5, with tr(A)=0, and p(A)=1, find the eigenvalues of A.
If the question would have been over C/R I would have found an answer, but this F field thing is too much for me...
Thanks!
Since the rank is 1, there exists vectors $\displaystyle x$ and $\displaystyle y$ such that $\displaystyle A = xy^T$(Prove this!)
This matrix has atmost one non zero eigenvalue. Clearly $\displaystyle x$ is an eigen vector corresponding to the non-zero(if it is...) eigenvalue $\displaystyle x^T y$.
But observe that $\displaystyle \text{Tr} (xy^T) = x^Ty$, thus all eigenvalues of A are 0.
Answer to second question:
Observe that $\displaystyle Ax = (xy^T)x = x(y^Tx) = (y^Tx)x$. If you still cant see it, observe that $\displaystyle y^Tx$ is a scalar.
Answer to question 1:
We will show that any eigenvector corresponding to a non-zero eigenvalue is a scalar multiple of x.
Let us say it had an eigenvector z, corresponding to a eigen value $\displaystyle \alpha$ ,then $\displaystyle Az = \alpha z \implies (xy^T)z = \alpha z \implies (y^T z) x = \alpha z $ , Thus if $\displaystyle \alpha \neq 0$, z is in the span of x.